Jesús de la Peña Hernández
n
x
L
l
180
cos
2
−
=
Therefore (1) is the necessary condition to twist till folding flat a stellate polygon. But this is
not sufficient: Besides we need that the paper will not interfere within itself. Something similar was
already treated in Point 8.2.8.5. Two conditions must be fulfilled: docility and availability; paper has to
be docile to flattening and has to be available without any interference.
Let ́s consider the aforesaid in connection with the pentagon. If we try to apply (1) to it we ́ll
come across Fig. 4: paper interference makes it impossible to construct a perfect stellate pentagon (the
small imperfection can be perceived).
Conversely, if we cheat as in Fig. 5, we make the construction possible: the folds take care of that little
error and an exempt stellate pentagon (Fig. 6) is obtained jutting out another convex pentagon.
11.4 HEXAGONAL STAR
VERSION 1 (H. Honda)
It is obtained from the big hexagon of Fig. 1.
Fig. 2 is the flattened hexagon with 2 a showing the two imbricated equilateral triangles
and 2 r, the hexagon in the back.
7
The construction made for the heptagon (x = 0,5
2
L
in
Fig.2, Point 11.2) was also a trick. One can compare that
value of x with x = 0,6235
2
L
which is obtained for (1) in
present Point 11.3. Fig. 7, reverse side, (Point 11.3) is the
result of using x = 0,6235
2
L
; the process takes care of the
error, but you can see the difference between Figs. 7 (Points
11.2 and 11.3).
(^45)
6